Help with Peig function

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Karin Norén-Cosgriff
Karin Norén-Cosgriff on 30 Aug 2018
Hi, I have problems understanding the meaning of nwin and nooverlaps in Peig. In the documentation it is stated that nwin is the window length and noverlap is the number of input sample points by which successive windows overlap. But the defualt values are 2xp(1) and nwin-1 respectively. Having a 18000 sample signal with one expected embedded sinusoid, I use p(1) = 2 in Peig. I would like to compare the results using a window length equal to the number of samples (and no everlaps), and dividing the signal in blocks with 6000 samples each and using 50 % overlap. How do I do this? Can someone explain and provide me with an example? Karin

Answers (1)

Vishal Bhutani
Vishal Bhutani on 3 Sep 2018
By my understanding you want to compare results of peig function using default values of nwin and noverlap. So you can find the sample example hope it will help:
fs = 100;
t = 0:1/fs:1-1/fs;
s = 2*sin(2*pi*25*t)+sin(2*pi*35*t)+randn(1,100);
X = corrmtx(s,12,'mod');
figure(1)
peig(X,2,512,fs) % default values of nwin = p(1) = 2 and noverlap = nwin-1.
figure(2)
peig(X,2,512,fs,20,10); % specifying nwin = 20, noverlap = 10.
For your case you can specify nwin = 6000 and noverlap = 3000.
  1 Comment
Karin Norén-Cosgriff
Karin Norén-Cosgriff on 4 Sep 2018
Thanks for you trying to expain this to me. However, I canot see that nwin and noverlap affects the results at all neither in your example, nor for my signal. Is it possible for you to show an example where this really has some affect for me to understand how they are working?

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